Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main()
{
	int n;
	cin>>n;
	long double a[n],v[n+1],prod=1,ans=0;
	for(int i=0;i<n;i++)
	{
		cin>>a[i];
		prod=prod*a[i];
	}
	for(int i=0;i<n+1;i++)
		v[i]=0;
	v[0]=1;
	for(int i=0;i<n;i++)
	{
		for(int j=i;j>=0;j--)
		{
			v[j+1]+=v[j]*(1-a[i])/a[i];
		}
	}
	for(int i=0;i<=n/2;i++)
	{
		ans=ans+prod*v[i];
	}
	cout.precision(18);
	cout<<fixed<<ans<<endl;
	return 0;
}